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Integration using long division

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.3 (EK)
Here we do polynomial long division to make an integral more computable.

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  • blobby green style avatar for user montgomery.tj
    I was working to find the integral of 3/(2x+4) using u-sub. If I factor out the 2 from the denominator and set u = x+2 I get the following solution: 3/2 * ln(x+2). However, if I do not factor and let u = 2x+4, I get 3/2 * ln(2x +4). What am I missing? Thanks.
    (20 votes)
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    • marcimus pink style avatar for user Alex Tran
      Remember that a general antiderivative of a function (indefinite integral) always has a constant of integration c attached to it. Assuming the above integral was done correctly, there should be a c attached to both. Notice that the first solution is 3/2 * ln(x+2) +c and the second is 3/2 * ln(2x+4) + c. Now manipulate (3/2) ln(2x+4) + c to get (3/2) ln(2*(x+2) ) + c and you get (3/2) ln(2) + (3/2) ln(x+2) +c by log properties. Notice now that 3/2 * ln(2) can be absorbed into the constant of integration, because it is a real number. Thus, we get (3/2) ln(x+2) + c for both the first and second solutions.

      Always remember to include the constant of integration.
      (38 votes)
  • female robot amelia style avatar for user micky.odunikan
    He did the long division as if this is something I should kinda know already is this covered in algebra or something?
    (10 votes)
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  • male robot donald style avatar for user harry park
    Hi, I need help solving a problem.
    The problem is to find the Anti-Derivative of x^2 + 2x - 2 + 15/(2x+3)
    The answer I came up with is : x^3/3 + x^2 - 2x + 15ln(2x+3) + C
    However, the correct answer is coming up as :
    :x^3/3 + x^2 - 2x + (15ln(2x+3))/2 + C
    My answer is almost identical to the correct answer except for the last part.
    I don't understand why the last expression is divided by 2.
    Can someone explain this to me please?
    Thank you.
    (4 votes)
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    • piceratops ultimate style avatar for user Just Keith
      Notice that in the standard integral forms, you ALWAYS have a du. This is NOT just some notation that you can ignore, it is vital to getting the correct answer. The du is the derivative of whatever you call u -- it MUST be present or you cannot use that standard derivative form.
      for the ∫ (1/u) du = ln(u) + C form, you must have the derivative of the denominator you are calling u.
      For15/(2x+3), you declared u to be 2x+3. That is fine, but you MUST have its derivative.
      u = 2x+3
      du = 2 dx
      in other words dx = ½ du
      So, to use this form I would need to do the following:
      ∫ 15 dx / (2x+3) ← though not necessary, factor out the 15 to avoid mistakes
      15∫ dx / (2x+3) ← now replace what you have with the u and du.
      15∫ (½ du) / (2x+3) ← we do this because dx = ½du
      15∫ (½ du) / u ← no we need to make this EXACTLY match the standard form.
      (15/2) ∫ du/ u ← I now have a standard form and can integrate
      (15/2) ln(u) + C ← now back substitute
      (15/2) ln(2x+3) + C
      Of course, we wouldn't typically show all of these steps, i just included them so you could get the idea.

      But the big thing to understand is this: In ALL of the standard integral forms there is a du. This is the derivative the whatever you call u. This MUST be present or you cannot use this form.

      For beginners, using u-substitution to make absolutely sure you have an exact match is the wisest course. Though, once you become proficient, you can probably skip the substitution and go straight to the integration. Here is the same calculation the way that I do it (this involves putting a bracket around my du, making sure I have the derivative of what I am calling u, although I won't actually use any substitutions:
      ∫ 15 dx / (2x+3) ← The "u" is 2x+3, so the "du" is 2dx.
      ∫ 15 (½) [2 dx] / (2x+3) ← notice that I multiplied and divided by 2, so I haven't changed anything.
      (15/2) ∫ [2 dx] / (2x+3) ← factored out the constants that are not in the standard integral form I am using.
      = (15/2) ln(2x+3) + C
      (9 votes)
  • piceratops seed style avatar for user Cagan Sevencan
    why would he use u substitution the point where he got 2/x-1 it's just seems unnecessary
    (3 votes)
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  • blobby green style avatar for user Jeron
    https://imgur.com/a/qA1ScKo

    In the above question for the integral of 1/(2x+6), if you factor out a 1/2 from the equation it becomes 1/2* integral of 1/(x+3) then doing u-sub you get 1/2*ln(x+3).

    How do you know when to factor out something versus not factoring something out because the 2 answers are different?
    (4 votes)
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    • leaf grey style avatar for user Alex
      It's technically the same thing.
      1 / 2 * ln(2x + 6) = 1 / 2 * [ln(2) + ln(x + 3)] = 1 / 2 * ln(x + 3) + C,
      where C = 1 / 2 * ln 2. The two integrals have a constant difference, and are therefore technically equivalent.
      (3 votes)
  • starky seedling style avatar for user George
    I am getting an extra +1/2 term compared with Sal's answer. I checked it with WolframAlpha and I get the same result as mine. Out of curiosity, I factored out from the denominator a (-2) and used u-sub (u = x - 1). I am trying to check for mistakes in my solution. Thanks for checking it out!
    (2 votes)
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  • piceratops tree style avatar for user Evan Phillips
    Can you factor out a (-2) from the denominator, changing the integral to (-1/2)*integral (x-5)/(x+1)? I get a different, but similar, answer: (-1/2)[x- 6 ln |x+1|] + C

    Appreciate the clarification!
    (2 votes)
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  • blobby green style avatar for user Haverworth Don
    In the video, "Integration using long division" the fraction 4/(2x-2) is simplified to 2*(1/(x-1)) to result in 2*ln(|x-1|). However, if you leave the fraction as 2*(2/(2x-2)) the result could be integrated as 2*ln(|2x-2|) which is a different function.

    1) is simplification before integration mandatory in these cases?
    2) If so, how do you avoid potentially changing a function when going to a derivative and back again?
    (2 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      ln|2𝑥 − 2| = ln|(𝑥 − 1) ∙ 2| = ln|𝑥 − 1| + ln 2.

      Thereby, ln|2𝑥 − 2| + 𝐶 is the same set of solutions as ln|𝑥 − 1| + 𝐶, and it doesn't really matter which representation we choose, it's just that (𝑥 − 1) is a little neater than (2𝑥 − 2), because it only involves one operation rather than two.
      (2 votes)
  • old spice man green style avatar for user Dania  Zaheer
    (1 vote)
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  • blobby green style avatar for user hcps-hubbartr1
    I don't know if someone asked this already, but what would you do if the denominator has a greater power than the numerator (like x^2 for the denominator and x for the numerator)
    (2 votes)
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    • female robot grace style avatar for user loumast17
      it still might work to do long division, there are also cases it may not work with an equal or lower degree for the denominator as well, in which case you just want to try other ways of simplifying or rewriting the numerator and denominator.
      (2 votes)

Video transcript

- [Voiceover] See if you can evaluate this integral right over here. Assuming you had a goal at it so let's work through this together. You probably realized that some of the traditional techniques that we've already had in our tool kits don't seem to be directly applicable, u-substitution and others. The key here to realize is we have a rational expression here where the numerator has the same degree or higher than the denominator. In this case, the numerator and the denominator have the same degree. Whenever you see something like that, it's probably a good idea to divide the denominator into the numerator. That's what this rational expression could be interpreted as. X minus five, divided by negative two x plus two. So it's a little bit of algebraic long division. To actually divide negative two x plus two into x minus five is to see if we can rewrite this in a way where we can evaluate the integral. So let's do that. We're going to take x minus five. X minus five, and divide negative two x plus two into that. Negative two x plus two. Look at the highest degree terms. How many times does negative two x go into x? Let's going to go negative 1/2 times. Negative 1/2 times two is negative one. Negative 1/2 times negative two x is this is going to be positive x. Just like that. Now we want to subtract this yellow expression from this blue expression, and so let's just, I'll just take the negative of this and then add. So I'm just going to take the negative of it and add. We are left with negative five plus one is negative four. You could say negative two x plus two goes into x minus five, negative 1/2 times with negative four left over. So we can rewrite this integral, our original integral as, we can rewrite it as negative 1/2 minus four over negative two x plus two, d x. Now let's see, looks like we can simplify this expression a little bit more. The numerator and the denominator, they're both divisible by two. All of these terms are divisible by two. Actually, we all have these negatives that always unnecessarily complicate things. Let's actually divide the numerator and the denominator by negative two. What are we going to have then? We divide the numerator by negative two. If this is negative four, this is going to become positive two. Then this, if we divide negative two x by negative two, that's just going to become x. Then two divided by negative two is going to be minus one. So our original integral... This is just all algebra. Everything we've done so far is algebra. We just rewritten it using a little bit of algebraic long division. Our original integral has simplified to negative 1/2, and some might argue it's not simplified but it's actually much more useful for finding the integral. Negative 1/2 plus two over x minus one, d x. Now, how do we evaluate this? Well, the antiderivative of negative 1/2 is pretty straight forward. That's just going to be negative 1/2 x, plus the entire derivative of two over x minus one. You might reduce here ahead. The derivative of x minus one is just one so you could say that the derivative is sitting there. We can essentially use substitution in our heads and say, "Okay, let's just take the entire derivative, "we could say with respect to x minus one which will be "the natural log of the absolute value of x minus one." If all of that sounds really confusing, I'll let you do the u-substitution. If I were just trying to evaluate the integral two over x minus one, d x, I could see, okay, the derivative of x minus one is just one so I could say u is equal to x minus one, and then d u is going to be equal to d x. This is going to be, we can rewrite in terms of u as two, I'll just take the constant out two times the integral of one over u, d u, which we know as two times the natural log of the absolute value of u plus c. In this case, we know that u is x minus one. This is equal to two times the natural log of x minus one plus c. That's what we're going to have right over here. So plus two times the natural log of the absolute value of x minus one plus c. The plus c doesn't just come from this one, this general overtaking the integral of the whole thing. It could be some constant because if we go the other way, we take the derivative and the constant will go away. Let me just put the plus c right over there, and we are done.